Question

Draw a $△ABC$ in which $BC=6cm,AB=4cm$ and $AC=5cm$. Draw a triangle similar to $△ABC$ with its sides equal to $(\frac{3}{4}{)}^{th}$ of the corresponding sides of $△ABC$.

Answer 1

The A' B' C' whose sides are $\frac{3}{4}$ of the corresponding sides of $\Delta ABC$ can be drawn as follows:-

Step 1 : Draw a $\Delta ABC$ with the following measurements :-

$BC=6cm,AB=5cm,\angle ABC={60}^o$

Step 2 : Draw a ray $BX$ make an acute angle with line $BC$ on the opposite of vertex A.

Step 3 : Locate 4 points that are $B_1,B_2,B_3,B_4$ on the $\overline{BX}$ (line segment BX)

Step 4 : Join $B_{4}C$ and draw a line through $B_3$ parallel to $B_{4}C$ intersecting BC at C'.

Step 5 : Draw a line through C' parallel to AC intersect AB at A'

$\Delta A^{\prime }{B}^{\prime }{C}^{\prime }$ is the below one :-

Ref. image